1288. More about side midpoints

    Given a triangle, its inellipses are its inscribed ellipses, that is, the ellipses tangent to the three sides of the triangle. The Steiner inellipse of the triangle is the unique inellipse passing through the three side midpoints of the triangle, that is, the ellipse tangent to the three sides at their midpoints. (Furthermore, the Steiner inellipse has the largest area of any inellipse.)

    Imagine a triangle on the complex plane... Marden's theorem states that if the three vertices of that triangle are the zeroes of a cubic polynomial, then the zeroes of the derivative of the polynomial are the foci of the Steiner inellipse... 
    Assuming Marden's theorem, can you prove that the center  (k)  of the Steiner inellipse, the barycenter  (g)  of the given triangle and the zero  (h)  of the second derivative of the cubic polynomial are the same point? Pepe Chapuzas was able...

    Dear Teacher:
    How beautiful is this theorem!
    If  a, b, c  are the vertices of the triangle, then its barycenter is

g  =  (a+b+c) / 3

    And the cubic polynomial and its derivatives are

P (z)  =  (z–a)·(z–b)·(z–c) · d     with    d ≠ 0

P (z)  =  ( z³ – (a+b+c)·z² + (ab+ac+bc)·z – abc ) · d

P ' (z)  =  ( 3·z² – 2·(a+b+c)·z + ab + ac + bc ) · d
P '' (z)  =  ( 6·z – 2·(a+b+c) ) · d

    According to Marden's theorem, the foci of the Steiner inellipse are the zeroes of  P ' (z) . If

3·z² – 2·(a+b+c)·z + ab + ac + bc  =  0

and  Δ Δ Δ is the discriminant of this equation, its zeroes (the foci) are

f₁  =  ( 2·(a+b+c) + √Δ ) / 6

f₂  =  ( 2·(a+b+c) – √Δ ) / 6

    The center of the Steiner inellipse shall be the midpoint of the segment between the foci:

k  =  ( f₁ + f₂ ) / 2  =  (a+b+c) / 3

    Finally, if  h  is the zero of  P '' (z)  then...

P '' (h)  =  ( 6·h – 2·(a+b+c) ) · d  =  0
6·h – 2·(a+b+c)  =  0

h  =  (a+b+c) / 3 

g  =  k  =  h    Q.E.D.